Question

A toy cannon uses a spring to project a 5.28-g soft rubber ball. The spring is originally compressed by 4.99 cm and has a force constant of 8.08 N/m. When the cannon is fired, the ball moves 15.2 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 2 N on the ball.

a)with what speed does the projectile leave the barrel of the cannon?
b)at what point does the ball have max speed?
c)What is the maximumspeed.

Answer 1

Step-by-step explanation:

When a spring is compressed , potential energy is stored in it which is released later on to impart kinetic energy to the fired ball.

If a spring is compressed by x , energy stored in it is

1/2 k x² where k is spring constant , x is compression.

In this case

potential energy stored

= 1/2 x 8.08 x (4.99)² x 10⁻⁴

= 100.6 x 10⁻⁴ J

If v be the velocity of ball just after firing

Kinetic energy of ball = 1/2 m v²

= .5 x 5.28 x 10⁻³ x v²

= 2.64 x 10⁻ ³ v²

According to law of conservation of energy

energy stored in spring = kinetic energy of ball

100.6 x 10⁻⁴ = 2.64 x 10⁻³ v²

v² = 38.11 x 10⁻¹

v = 1.95 m /s

During travel through barrel , the velocity of ball is reduced due to work done by friction.

Work done by friction

Friction force x distance

.032 x 0.152 J

= 4.89 X 10⁻³ J

Loss of energy of ball

= 4.89 X 10⁻³ J

Net energy of the ball while coming out of barrel

= 100.6 x 10⁻⁴ - 4.89 X 10⁻³

= 10.06 x 10⁻³ - 4.89 x 10⁻³

= 5.17 x 10⁻³

If v₁ be the velocity of emergence of ball out of barrel

1/2 m v₁² = 5.17 x 10⁻³

.5 x 5.28 x 10⁻³ v₁² = 5.17 x 10⁻³

v₁² = .5106

v₁ = .7145 m/s

b ) The maximum speed will be at a point of time when spring un- compresses completely to impart velocity to the ball.

c ) The maximum speed will be equal to

1.95 m /s